Engineer's Notebook §4.1

Constant acceleration

When acceleration is constant, speed, distance and time are tied together by the equations of motion. Without needing the time:

v² = u² + 2as
u
initial speed
m/s
v
final speed
m/s
a
acceleration
m/s²
s
displacement
m
t
time
s
A speed–time graph. A straight line runs from the initial speed to the final speed, and the shaded area beneath it is the distance travelled. TIME tSPEED v u = 5 v = 15 m/s t = 5 s s = 50 m
Fig. 4.1-1The area under the line is the distance travelled. Axes run to 20 s and 40 m/s.
t = (v − u)/a5 s
(u + v)/2, mean speed10 m/s
v = √(u² + 2as)15 m/s

Worked example

Uses the values in Fig. 4.1-1. Change them and the working updates.

Given
  • uinitial speed5 m/s
  • aconstant acceleration2 m/s²
  • sdistance travelled50 m
Find

The final speed v and the time taken t.

Solution
v² = u² + 2as = (5 m/s)² + 2 × 2 m/s² × 50 m = 225 m²/s² v = √225 = 15 m/s t = (v − u)/a = (15 − 5) m/s / 2 m/s² = 5 s
Answer
v = 15 m/s

That is 54 km/h, reached after 5 s.

The other equations of motion

  • Speed after time t

    v = u + at

  • Distance from time

    s = ut + ½at²

  • Distance from mean speed

    s = ½(u + v)t

  • Stopping distance from u

    s = u²2a

Assumptions
  • Acceleration is constant for the whole motion.
  • Motion is along a straight line, and u, v, a and s are signed along it.
  • Drag and other speed-dependent forces are ignored, or folded into a constant a.

Derivation
  1. Acceleration is the slope of the speed–time graph: v = u + at.
  2. Distance is the area under it, a trapezium: s = (u + v)t/2.
  3. Take t = (v − u)/a from the first and substitute: s = (u + v)(v − u)/2a = (v² − u²)/2a.
  4. Rearrange: v² = u² + 2as.